2つのフォーミュラ1認定フォーマットの統計的変動

フォーミュラ1の予選フォーマットに関するこのBBCの記事を読んだばかりです。

主催者は、予選の予測可能性を低くしたい、つまり結果の統計的変動を増やしたいと考えています。いくつかの関連性のない詳細を明らかにし、現時点では、ドライバーは（具体性のために）2回の試行のベストシングルラップでランク付けされています。

F1チーフの1人であるジャントッドは、ドライバーがミスを犯す可能性が2倍になる可能性があるため、ドライバーを2ラップの平均でランク付けすると統計的変動が増えると提案しました。他の情報源は、平均化は統計的変動を確実に減少させると主張した。

合理的な仮定の下で誰が正しいと言えますか？$\text{mean}(x,y)$対の相対分散にとます。ここで、とはドライバーの2つのラップタイムを表すランダム変数ですか？$\text{min}(x,y)$$x$$y$

ラップタイムの分布に依存すると思います。

ましょ同一分布、独立しています。$X,Y$

1. もし、次にVar（X+$P(X=0)=P(X=1)=\frac{1}{2}$$Var(\frac{X+Y}{2}) = \frac{1}{8} < Var( \min(X,Y)) = \frac{3}{16}.$
2. ただし、場合、 V a r （X + $P(X=0) = 0.9, P(X=100)=0.1$$Var(\frac{X+Y}{2}) = 450 > Var( \min(X,Y)) = 99.$

これは、間違いを犯すことに関する質問で言及された議論（つまり、非常に長い時間をわずかな確率で実行する）と一致しています。したがって、決定するにはラップタイムの分布を知る必要があります。

一般性を失うことなく、想定との両方varaiblesが特定の平均と分散と同一の分布から引き出されていること。$y \leq x$

は { x }を改善し、$\{y,x\}$$\{x\}$

ケース1、平均： 、$\frac{y-x}{2}$

ケース2分：。$y-x$

したがって、平均は、最小値をとる（2回の試行の）場合の改善（分散によって駆動される）の半分の効果を持ちます。つまり、平均は変動性を減衰させます。

これが私のVarの証明です[平均]

2つのランダム変数x、yの場合、それらの平均値と最大値および最小値の間には関係があります。

したがって

$$2\,Mean(x,y) = Min(x,y) + Max(x,y)$$ $$4\,Var[Mean] = Var[Min]+Var[Max]+2\,Cov[Min,Max]$$ $$Var[Min(x,y)]=Var[Max(x,y)]$$ $$4\,Var[Mean] = 2\,Var[Min] + 2\,Cov[Min,Max]$$ and $$Cov[Min,Max]<=sqrt(Var[Min]Var[Max])=Var[Min]$$ Therefore $$Var[Mean]<=Var[Min]$$ It is also easy to see from this derivation that in order to reverse this inequality you need a distribution with very sharp truncation of the distribution on the negative side of the mean. For example for the exponential distribution the mean has a larger variance than the min.

Nice question, thank you! I agree with @sandris that distribution of lap times matters, but would like to emphasize that causal aspects of the question need to be addressed. My guess is that F1 wants to avoid a boring situation where the same team or driver dominates the sport year after year, and that they especially hope to introduce the (revenue-generating!) excitement of a real possibility that 'hot' new drivers can suddenly arise in the sport.

That is, my guess is that there is some hope to disrupt excessively stable rankings of teams/drivers. (Consider the analogy with raising the temperature in simulated annealing.) The question then becomes, what are the causal factors at work, and how are they distributed across the population of drivers/teams so as to create persistent advantage for current incumbents. (Consider the analogous question of levying high inheritance taxes to 'level the playing field' in society at large.)

Suppose incumbent teams are maintaining incumbency by a conservative strategy heavily dependent on driver experience, that emphasizes low variance in lap times at the expense of mean lap time. Suppose that by contrast the up-and-coming teams with (say) younger drivers, necessarily adopt a more aggressive (high-risk) strategy with larger variance, but that this involves some spectacular driving that sometimes 'hits it just right' and achieves a stunning lap time. Abstracting away from safety concerns, F1 would clearly like to see some such 'underdogs' in the race. In this causal scenario, it would seem that a best-of-n-laps policy (large $n$) would help give the upstarts a boost -- assuming that the experienced drivers are 'set in their ways', and so couldn't readily adapt their style to the new policy.

Suppose, on the other hand, that engine failure is an uncontrollable event with the same probability across all teams, and that the current rankings correctly reflect genuine gradation in driver/team quality across many other factors. In this case, the bad luck of an engine failure promises to be the lone 'leveling factor' that F1 could exploit to achieve greater equality of opportunity--at least without heavy-handed ranking manipulations that destroy the appearance of 'competition'. In this case, a policy that heavily penalizes engine failures (which are the only factor in this scenario not operating relatively in favor of the incumbents) promises to promote instability in rankings. In this case, the best-of-n policy mentioned above would be exactly the wrong policy to pursue.

I generally agree with other answers that the average of two runs will have a lower variance, but I believe they are leaving out important aspects underlying the problem. A lot has to do with how drivers react to the rules and their strategies for qualifying.

For instance, with only one lap to qualify, drivers would be more conservative, and therefore more predictable and more boring to watch. The idea with two laps is to allow the drivers to take chances on one to try to get that "perfect lap", with another available for a conservative run. More runs would use up a lot of time, which could also be boring. The current setup might just be the "sweet spot" to get the most action in the shortest time frame.

Also note that with an averaging approach, the driver needs to find the fastest repeatable lap time. With the min approach, the driver needs to drive as fast as possible for only one lap, probably pushing further than they would under the averaging approach.

This discussion is closer to game theory. Your question might get better answers when framed in that light. Then one could propose other techniques, like the option for a driver to drop the first lap time in favor of a second run, and possibly a faster or slower time. Etc.

Also note that a change in qualifying was attempted this year that generally pushed drivers into one conservative lap. https://en.wikipedia.org/wiki/2016_Formula_One_season#Qualifying The result was viewed as a disaster and quickly cancelled.